Problem: Drazil likes heap very much. So he created a problem with heap:

There is a max heap with a height h implemented on the array. The details of this heap are the following:

This heap contains exactly 2^h - 1 distinct positive non-zero integers. All integers are distinct. These numbers are stored in the array a indexed from 1 to 2^h-1. For any 1 < i < 2^h, a[i] < a[\left ⌊{i/2}\right ⌋].

Now we want to reduce the height of this heap such that the height becomes g with exactly 2^g-1 numbers in heap. To reduce the height, we should perform the following action 2^h-2^g times:

Choose an index i, which contains an element and call the following function f in index i:

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Note that we suppose that if a[i]=0, then index i don't contain an element.

After all operations, the remaining 2^g-1 element must be located in indices from 1 to 2^g-1. Now Drazil wonders what's the minimum possible sum of the remaining 2^g-1 elements. Please find this sum and find a sequence of the function calls to achieve this value.

Input

The first line of the input contains an integer t (1 ≤ t ≤ 70 000): the number of test cases.

Each test case contain two lines. The first line contains two integers h and g (1 ≤ g < h ≤ 20). The second line contains n = 2^h-1 distinct positive integers a[1], a[2], …, a[n] (1 ≤ a[i] < 2^{20}). For all i from 2 to 2^h - 1, a[i] < a[\left ⌊{i/2}\right ⌋].

The total sum of n is less than 2^{20}.

Output

For each test case, print two lines.

The first line should contain one integer denoting the minimum sum after reducing the height of heap to g. The second line should contain 2^h - 2^g integers v_1, v_2, …, v_{2^h-2^g}. In i-th operation f(v_i) should be called.

Example

Input


2
3 2
7 6 3 5 4 2 1
3 2
7 6 5 4 3 2 1


Output


10
3 2 3 1
8
2 1 3 1

Solution (in JAVA):

Solution:
import java.util.*;
import java.io.*;
import java.text.*;

public class Main {
    //SOLUTION BEGIN
    //Into the Hardware Mode
    void pre() throws Exception{}
    void solve(int TC) throws Exception{
        int H = ni(), G = ni();
        int N = (1<<H)-1;
        int[] A = new int[1+N], dep = new int[1+N];
        int[] invA = new int[1+N];
        int[] B = new int[1+N];
        for(int i = 1; i<= N; i++){
            A[i] = ni();
            B[i] = A[i];
            dep[i] = dep[i/2]+1;
        }
        Arrays.sort(B);
        for(int i = 1; i<= N; i++)invA[A[i] = Arrays.binarySearch(B, A[i])] = i;
        int[] ans = new int[(1<<H)-(1<<G)];
        for(int i = N, j = 0; i >= 1; i--){
            if(!canDelete(A, dep, invA[i], G))continue;
            ans[j++] = invA[i];
            delete(A, dep, invA, invA[i]);
            invA[i] = -1;
        }
        long sum = 0;
        for(int i = 1; i<= N; i++)sum += B[A[i]];
        pn(sum);
        for(int i = 0; i< ans.length; i++)p(ans[i]+" ");pn("");
    }
    boolean canDelete(int[] A, int[] dep, int pos, int G){
        if(dep[pos] > G)return true;
        int le = pos<<1, ri = pos<<1|1;
        if(A[le] == 0 && A[ri] == 0)return false;
        if(A[le] > A[ri])
            return canDelete(A, dep, pos<<1, G);
        else
            return canDelete(A, dep, pos<<1|1, G);
    }
    void delete(int[] A, int[] dep, int[] invA, int pos){
        int le = pos<<1, ri = pos<<1|1;
        if(ri > dep.length)
            A[pos] = 0;
        else{
            if(A[le] > A[ri]){
                invA[A[le]] = pos;
                A[pos] = A[le];
                delete(A, dep, invA, le);
            }else{
                invA[A[ri]] = pos;
                A[pos] = A[ri];
                delete(A, dep, invA, ri);
            }
        }
    }
    //SOLUTION END
    void hold(boolean b)throws Exception{if(!b)throw new Exception("Hold right there, Sparky!");}
    void exit(boolean b){if(!b)System.exit(0);}
    static void debug(Object... o){System.err.println(Arrays.deepToString(o));}
    final long IINF = (long)2e18;
    final int INF = (int)1e9+2;
    DecimalFormat df = new DecimalFormat("0.00000000000");
    double PI = 3.141592653589793238462643383279502884197169399, eps = 1e-8;
    static boolean multipleTC = true, memory = false, fileIO = false;
    FastReader in;PrintWriter out;
    void run() throws Exception{
        long ct = System.currentTimeMillis();
        if (fileIO) {
            in = new FastReader("");
            out = new PrintWriter("");
        } else {
            in = new FastReader();
            out = new PrintWriter(System.out);
        }
        //Solution Credits: Taranpreet Singh
        int T = multipleTC? ni():1;
        pre();
        for (int t = 1; t <= T; t++) solve(t);
        out.flush();
        out.close();
        System.err.println(System.currentTimeMillis() - ct);
    }
    public static void main(String[] args) throws Exception{
        if(memory)new Thread(null, new Runnable() {public void run(){try{new Main().run();}catch(Exception e){e.printStackTrace();}}}, "1", 1 << 28).start();
        else new Main().run();
    }
    int[][] make(int n, int[] from, int[] to, int e, boolean f){
        int[][] g = new int[n][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = to[i];
            if(f)g[to[i]][--cnt[to[i]]] = from[i];
        }
        return g;
    }
    int[][][] makeS(int n, int[] from, int[] to, int e, boolean f){
        int[][][] g = new int[n][][];int[]cnt = new int[n];
        for(int i = 0; i< e; i++){
            cnt[from[i]]++;
            if(f)cnt[to[i]]++;
        }
        for(int i = 0; i< n; i++)g[i] = new int[cnt[i]][];
        for(int i = 0; i< e; i++){
            g[from[i]][--cnt[from[i]]] = new int[]{to[i], i, 0};
            if(f)g[to[i]][--cnt[to[i]]] = new int[]{from[i], i, 1};
        }
        return g;
    }
    int find(int[] set, int u){return set[u] = (set[u] == u?u:find(set, set[u]));}
    int digit(long s){int ans = 0;while(s>0){s/=10;ans++;}return ans;}
    long gcd(long a, long b){return (b==0)?a:gcd(b,a%b);}
    int gcd(int a, int b){return (b==0)?a:gcd(b,a%b);}
    int bit(long n){return (n==0)?0:(1+bit(n&(n-1)));}
    void p(Object o){out.print(o);}
    void pn(Object o){out.println(o);}
    void pni(Object o){out.println(o);out.flush();}
    String n()throws Exception{return in.next();}
    String nln()throws Exception{return in.nextLine();}
    int ni()throws Exception{return Integer.parseInt(in.next());}
    long nl()throws Exception{return Long.parseLong(in.next());}
    double nd()throws Exception{return Double.parseDouble(in.next());}

    class FastReader{
        BufferedReader br;
        StringTokenizer st;
        public FastReader(){
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        public FastReader(String s) throws Exception{
            br = new BufferedReader(new FileReader(s));
        }

        String next() throws Exception{
            while (st == null || !st.hasMoreElements()){
                try{
                    st = new StringTokenizer(br.readLine());
                }catch (IOException  e){
                    throw new Exception(e.toString());
                }
            }
            return st.nextToken();
        }

        String nextLine() throws Exception{
            String str;
            try{
                str = br.readLine();
            }catch (IOException e){
                throw new Exception(e.toString());
            }
            return str;
        }
    }
}